Use the quadratic formula to find the solution(s) of $ \frac{1}{2}x^2 - 3x + 4 = 0 $. $ x = 2 $ none \(...

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2. Select all that apply. Use the quadratic formula to find the solution(s) of $ \frac{1}{2}x^2 - 3x + 4 = 0 $. $ x = 2 $ none $ x = 4 $ $ x = -4 $ $ x = -2 $

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To solve $ \frac{1}{2}x^2 - 3x + 4 = 0 $ using the quadratic formula, we first identify $ a = \frac{1}{2} $, $ b = -3 $, and $ c = 4 $.The quadratic formula is $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $.Calculate the discriminant: $ b^2 - 4ac = (-3)^2 - 4 \times \frac{1}{2} \times 4 = 9 - 8 = 1 $.Since the discriminant is positive, there are two real solutions: $ x = \frac{-(-3) + \sqrt{1}}{2 \times \frac{1}{2}} = \frac{3 + 1}{1} = 4 $ $ x = \frac{-(-3) - \sqrt{1}}{2 \times \frac{1}{2}} = \frac{3 - 1}{1} = 2 $ Therefore, the solutions are $ x = 4 $ and $ x = 2 $.Correct options: $ x = 2 $, $ x = 4 $

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Use the quadratic formula to find the solution(s) of \( \frac{1}{2}x^2 - 3x + 4 = 0 \). \( x = 2 \) none \(...

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